Find the volume of the solid bounded by the sphere
Find the volume of the solid bounded by the sphere: $$ x^2+y^2+z^2=3a^2 $$ and the paraboloid: $$ x^2+y^2=2az, $$ with $z>0$ taking the region containing the smaller of the two pieces cut off by the paraboloid (i.e. between the paraboloid and the sphere).
Explanation of Method
A note on this problem: this is the same sphere/paraboloid configuration as Problem 160 — identical surfaces, identical setup, identical answer. The solution is repeated here in full for completeness.
The sphere and paraboloid intersect where $2az+z^2=3a^2$, i.e. $z=a$ (taking the positive root). In polar coordinates, the paraboloid gives $z=\frac{r^2}{2a}$ and the sphere gives $z=\sqrt{3a^2-r^2}$; the volume between them is found by integrating the height difference over the disk of intersection.
- Find the intersection height $z=a$ (radius $r=a\sqrt2$).
- Set up the volume as the sphere-cap volume plus the region under the sphere down to the paraboloid, in polar coordinates.
- Evaluate the resulting integral.
Step 1 · Find the intersection
Substituting $x^2+y^2=2az$ into the sphere's equation: $2az+z^2=3a^2\Rightarrow z^2+2az-3a^2=0\Rightarrow(z-a)(z+3a)=0$. Taking $z>0$: $z=a$, at radius $r^2=2a\cdot a=2a^2$, i.e. $r=a\sqrt2$.
Step 2 · Set up the volume in polar coordinates
The solid bounded below by the paraboloid and above by the sphere, over the disk $r\leqslant a\sqrt2$:
$$ V=\int_0^{2\pi}\int_0^{a\sqrt2}\left(\sqrt{3a^2-r^2}-\frac{r^2}{2a}\right)r\,dr\,d\theta. $$
Step 3 · Evaluate
$$ \int_0^{a\sqrt2}\sqrt{3a^2-r^2}\,r\,dr = \left[-\frac{(3a^2-r^2)^{3/2}}3\right]_0^{a\sqrt2} = \frac{(3a^2)^{3/2}-a^3}3 = \frac{3\sqrt3\,a^3-a^3}3, $$
$$ \int_0^{a\sqrt2}\frac{r^3}{2a}\,dr = \frac1{2a}\cdot\frac{(a\sqrt2)^4}4 = \frac{4a^4}{8a}=\frac{a^3}2. $$
So the radial integral is $\dfrac{3\sqrt3\,a^3-a^3}3-\dfrac{a^3}2 = a^3\left(\sqrt3-\frac13-\frac12\right)=a^3\left(\sqrt3-\frac56\right)$, and:
$$ V=2\pi\,a^3\left(\sqrt3-\frac56\right) = \frac{\pi a^3\big(6\sqrt3-5\big)}3. $$
Additional Notes
- This problem is identical in setup and result to Problem 160.
- Verified symbolically: direct evaluation of the polar double integral gives exactly $\dfrac{\pi a^3(6\sqrt3-5)}3$.
Result
The volume is: